Cần lắm hông

a/ \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(3-2\sqrt{5}\right)^2}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=1\)

b,c tương tự

a) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

\(=\sqrt{5-\sqrt{3-\sqrt{20-2\cdot3\cdot\sqrt{20}+9}}}\)

\(=\sqrt{5-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)

\(=\sqrt{5-\sqrt{3-\sqrt{20}+3}}\)

\(=\sqrt{5-\sqrt{6-\sqrt{20}}}\)

\(=\sqrt{5-\sqrt{5-2\sqrt{5}+1}}\)

\(=\sqrt{5-\sqrt{\left(\sqrt{5}+1\right)^2}}\)

\(=\sqrt{5-\sqrt{5}-1}\)

\(=\sqrt{4-\sqrt{5}}\)

c)\(\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\)

\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{3+2\cdot\sqrt{3}\cdot\sqrt{2}+2}\)

\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)

\(=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)\)

\(=3-2=1\)

d)\(\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\)

\(=\sqrt{5-\sqrt{12+2\cdot\sqrt{12}+1}}+\sqrt{3+\sqrt{12+2\cdot\sqrt{12}+1}}\)

\(=\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}+\sqrt{3+\sqrt{\left(\sqrt{12}+1\right)^2}}\)

\(=\sqrt{5-\sqrt{12}-1}+\sqrt{3+\sqrt{12}+1}\)

\(=\sqrt{4-\sqrt{12}}+\sqrt{4+\sqrt{12}}\)

\(=\sqrt{3-2\sqrt{3}+1}+\sqrt{4+2\sqrt{3}+1}\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(=\sqrt{3}-1+\sqrt{3+1}\)

\(=2\sqrt{3}\)

\(\sqrt{24+8\sqrt{5}}+\) \(\sqrt{9-4\sqrt{5}}=\) \(\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}.2+4}\) + \(\sqrt{5-2\sqrt{5}.2+4}\)

= \(\sqrt{\left(2\sqrt{5}+2\right)^2}+\) \(\sqrt{\left(\sqrt{5}-2\right)^2}\) = \(2\sqrt{5}+2+\sqrt{5}-2=3\sqrt{5}\)

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\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\) = \(\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)= \(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)

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\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)

= \(\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\) \(=\sqrt{\left(3\sqrt{2}+5\right)^2}=3\sqrt{2}+5\)

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c/ = \(\sqrt{13+30\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}\)

\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\)

\(=\sqrt{43+30\sqrt{2}}\)

\(=\sqrt{25+2.3.5.\sqrt{2}+18}\)

\(=5+3\sqrt{2}\)

d/ \(=\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}\)

\(=\sqrt{13+6\sqrt{4+2\sqrt{2}-1}}\)

\(=\sqrt{13+6\left(\sqrt{3}+1\right)}\)

\(=\sqrt{19+6\sqrt{2}}\)

\(=3\sqrt{2}+1\)

Lời giải:

a. \(\sqrt{6-2\sqrt{5}}=\sqrt{5-2\sqrt{5}.\sqrt{1}+1}=\sqrt{(\sqrt{5}-1)^2}=\sqrt{5}-1\)

b. \(\sqrt{7-4\sqrt{3}}=\sqrt{4-2\sqrt{4}.\sqrt{3}+3}=\sqrt{(\sqrt{4}-\sqrt{3})^2}=\sqrt{4}-\sqrt{3}=2-\sqrt{3}\)

c.

\(\sqrt{3-2\sqrt{2}}-\sqrt{6-4\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}-\sqrt{4-4\sqrt{2}+2}\)

\(=\sqrt{(\sqrt{2}-1)^2}-\sqrt{(\sqrt{4}-\sqrt{2})^2}\)

\(=|\sqrt{2}-1|-|\sqrt{4}-\sqrt{2}|=\sqrt{2}-1-(2-\sqrt{2})=2\sqrt{2}-3\)

d.

\(=\sqrt{13+30\sqrt{2+\sqrt{(\sqrt{8}+1)^2}}}=\sqrt{13+30\sqrt{2+\sqrt{8}+1}}\)

\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\sqrt{(\sqrt{2}+1)^2}}\)

\(=\sqrt{13+30(\sqrt{2}+1)}=\sqrt{43+30\sqrt{2}}=\sqrt{18+2\sqrt{18.25}+25}\)

\(=\sqrt{(\sqrt{18}+\sqrt{25})^2}=\sqrt{18}+\sqrt{25}=5+3\sqrt{2}\)

a) \(\sqrt{6-2\sqrt{5}}=\sqrt{5}-1\)

b) \(\sqrt{7-4\sqrt{3}}=2-\sqrt{3}\)

c) \(\sqrt{3-2\sqrt{2}}-\sqrt{6-4\sqrt{2}}=\sqrt{2}-1-2+\sqrt{2}=-3+2\sqrt{2}\)

d) Ta có: \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

\(=\sqrt{13+30\sqrt{2+1+2\sqrt{2}}}\)

\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)

\(=\sqrt{43+30\sqrt{2}}\)

\(=5+3\sqrt{2}\)

\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2.2\sqrt{5}.3+9}}}\)

 \(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)

\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{5-2.\sqrt{5}.1+1}}\)

\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)

\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)

\(=\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)

\(\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}\)

tương tự như trên 

\(=\sqrt{6+2\sqrt{5}-\left(2\sqrt{5}-3\right)}\)

\(=\sqrt{6+2\sqrt{5}-2\sqrt{5}+3}=\sqrt{9}=3\)

chúc bn học tốt

ò, mình hiểu cách làm của bạn rồi, nhưng mà mình nghĩ chỗ câu a), câu b) bạn giải chỗ dấu ''='' thứ 3, sau khi nhận dạng đó là \(\sqrt{A^2}=|A|\), thì bạn phải bằng ra trị căn A, rồi nếu đó là phép cộng thì viết thẳng ra,còn nếu phép trừ thì phải xét xem là A nhỏ hơn 0 thì trị A= - A, còn nếu lớn hơn hoặc bằng 0 thì bằng chính nó,  đồng ý với bạn là ngoài là dấu trừ nên để trong ngoặc nhưng làm như vậy thì gọi là bỏ bước nếu bạn là hsg thì mình không có ý kiến nhưng mà bạn bỏ cái bước trị tuyệt dối nhưng lại không bỏ bước đặt dấu ngoặc, làm vậy cũng đúng nếu bạn không vững quy tắc dấu, nhưng mà cái bước trị tuyệt đối quan trọng hơn. Mình nghĩ vậy!

Còn câu c) bạn làm ra kết quả sao? 

a) Ta có: \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2\cdot\sqrt{20}\cdot3+9}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)

\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{5-2\cdot\sqrt{5}\cdot1+1}}\)

\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)

\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)

\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)

\(=\sqrt{1}=1\)

b) Ta có: \(\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}\)

\(=\sqrt{6+2\sqrt{5}-\sqrt{20-2\cdot2\sqrt{5}\cdot3+9}}\)

\(=\sqrt{6+2\sqrt{5}-\sqrt{\left(2\sqrt{5}-3\right)^2}}\)

\(=\sqrt{6+2\sqrt{5}-\left(2\sqrt{5}-3\right)}\)

\(=\sqrt{6+3}=3\)

c) Sửa đề: \(\sqrt{2+\sqrt{5+\sqrt{13-\sqrt{48}}}}\)

Ta có: \(\sqrt{2+\sqrt{5+\sqrt{13-\sqrt{48}}}}\)

\(=\sqrt{2+\sqrt{5+\sqrt{12-2\cdot2\sqrt{3}\cdot1+1}}}\)

\(=\sqrt{2+\sqrt{5+\sqrt{\left(2\sqrt{3}-1\right)^2}}}\)

\(=\sqrt{2+\sqrt{5+2\sqrt{3}-1}}\)

\(=\sqrt{2+\sqrt{3+2\sqrt{3}\cdot1+1}}\)

\(=\sqrt{2+\sqrt{\left(\sqrt{3}+1\right)^2}}\)

\(=\sqrt{3+\sqrt{3}}\)

d) Ta có: \(\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)

\(=\dfrac{\left(6-2\sqrt{5}\right)\sqrt{6+2\sqrt{5}}+\left(6+2\sqrt{5}\right)\sqrt{6-2\sqrt{5}}}{2\sqrt{2}}\)

\(=\dfrac{\left(\sqrt{5}-1\right)^2\cdot\left(\sqrt{5}+1\right)+\left(\sqrt{5}+1\right)^2\cdot\left(\sqrt{5}-1\right)}{2\sqrt{2}}\)

\(=\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\left(\sqrt{5}-1+\sqrt{5}+1\right)}{2\sqrt{2}}\)

\(=\dfrac{4\cdot2\sqrt{5}}{2\sqrt{2}}\)

\(=\dfrac{8\sqrt{5}}{2\sqrt{2}}=\dfrac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)

\(\sqrt{13+30\sqrt{2}+\sqrt{9+4\sqrt{2}}}\)

\(=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}\right)^2}+4\sqrt{2}+1^2}}\)

\(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1^2}}\)

\(=\sqrt{13+30\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}+1^2}}\)

\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)

\(=\sqrt{5^2+2.5.3\sqrt{2}+\left(3+\sqrt{2}\right)^2}\)

\(=\sqrt{\left(5+3+\sqrt{2}\right)^2}\)

\(=\sqrt{\left(5+6\right)}=\sqrt{11}\)

\(=5+6=11\)

a)=1
b)=\(3\sqrt{2}+5\)

a)  \(\sqrt{\sqrt{5}-\sqrt{9-\sqrt{29-12\sqrt{5}}}}\)

=\(\sqrt{\sqrt{5}-\sqrt{9-\sqrt{(2\sqrt{5}+3)^2}}}\)

=\(\sqrt{\sqrt{5}-\sqrt{9-(2\sqrt{5}+3)}}\)

=\(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)

=\(\sqrt{\sqrt{5}-(\sqrt{5}-1)}\)

=\(\sqrt{1}=1\)

b)  \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

=\(\sqrt{13+30\sqrt{2+\sqrt{(2\sqrt{2}+1)^2}}}\)

=\(\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)

=\(\sqrt{13+30\sqrt{(2+1)^2}}\)

=\(\sqrt{13+30.(2+1)}\)

=\(\sqrt{43+30\sqrt{2}}=\sqrt{(3\sqrt{2}+5)^2}\)

=\(3\sqrt{2}+5\)